In what way is the calculation of $\lim_{x \to \infty} \frac{4x^2}{x-2}$ wrong? Is something wrong with the calculation below?
$$ \begin{align}
\lim_{x \to \infty} \frac{4x^2}{x-2} &= \lim_{x \to \infty} \frac{4x}{1-2/x} \\
& = \frac{(\lim_{x \to \infty} 4x)}{(\lim_{x \to \infty} 1-2/x)} \\
& = \frac{(\lim_{x \to \infty} 4x)}{1} \\
& = \lim_{x \to \infty} 4x \\
& = \infty.
\end{align} $$
I ask because if there isn't then the following would seem correct,
$$ \begin{align}
\lim_{x \to \infty} \frac{4x^2}{x-2} - 4x &= \left( \lim_{x \to \infty} \frac{4x^2}{x-2} \right) - \left( \lim_{x \to \infty} 4x \right) \\
& = \left( \lim_{x \to \infty} 4x \right) - \left( \lim_{x \to \infty} 4x \right) \\
& = \lim_{x \to \infty} 4x - 4x \\
& = \lim_{x \to \infty} 0 \\
& = 0.
\end{align} $$
But it is not, since
$$ \begin{align}
\lim_{x \to \infty} \frac{4x^2}{x-2} - 4x &= \lim_{x \to \infty} \frac{4x^2 - 4x(x-2)}{x-2} \\
&= \lim_{x \to \infty} \frac{8x}{x-2} \\
&= \lim_{x \to \infty} \frac{8}{1-2/x} \\
&= 8. \\
\end{align} $$
In what way is this wrong? Where is the mistake?
 A: $\infty-\infty$ is an indeterminate form, so the result of it can be anything.
Your mistake is exactly here:
$$\left( \lim_{x \to \infty} 4x \right) - \left( \lim_{x \to \infty} 4x \right)= \lim_{x \to \infty} (4x - 4x)$$
On the left side you have $\infty-\infty$, which has to be solved on a case by case basis by returning to the original limit. So splitting in this case does not work.
A: You can't split a limit term-by-term like that unless both terms converge.  In this case, neither does, so you can't.
A: It is incorrect that

$$
\lim_{x \to \infty} \frac{4x^2}{x-2} - 4x = \left( \lim_{x \to \infty} \frac{4x^2}{x-2} \right) - \left( \lim_{x \to \infty} 4x \right)
$$

since in order to apply the limit rule
$$
\lim_{x\to\infty}\big(f(x)-g(x)\big)=\lim_{x\to\infty}f(x)-\lim_{x\to\infty}g(x)
$$
one needs both at least one of $\lim_{x\to\infty}f(x)$ and $\lim_{x\to\infty}g(x)$ being real numbers. ([Added later:] In the case $\infty-(-\infty)$, one needs to define the arithmatic operation for the extended real numbers first.)

[Added:]
It is dangerous to view $\infty$ as a number unless you know exactly what extended real numbers are and what arithmetic operations are (and are not) allowed for them. There are more examples for nonsense by considering $\infty$ as a real number:
$$
\lim_{x\to\infty}\left(x\cdot\frac{1}{x}\right)=\infty\cdot 0=0;
$$
$$
\lim_{x\to\infty}\big(2x-x\big)=\infty-\infty=0
$$

[Added later:]
Also, the step
$$ \lim_{x \to \infty} \frac{4x}{1-2/x} 
= \frac{(\lim_{x \to \infty} 4x)}{(\lim_{x \to \infty} 1-2/x)}
$$
would not be mathematically correct unless one is working in the extended real numbers and has defined what is $\dfrac{\infty}{a}$ for a non-zero real number $a$.
